Question

Approximate the value of ^3 sqrt 27.3 using the tangent line of f(x)= ^3 sqrt x at point x= 27?

Answer 1

The point of tangency will be `f(27) = root(3)(27) = 3` so `(27, 3)`.

The slope of the tangent at this point is given by the derivative.

`f(x) = x^(1/3) -> f'(x) = 1/3x^(-2/3)`

`f'(27) = 1/3(27)^(-2/3) = 1/(3(27)^(2/3)) = 1/27`

Recall the equation of a line is given by

`y -y_1 = m(x- x_1)`

`y - 3 = 1/27(x - 27)`

`y = 1/27x - 1 +3`

`y = 1/27x + 2`

So at `27.3`, the tangent line approximation gives us

`y(27.3) = 1/27(27.3) + 2 = 3.011`

If you check using a calculator, your answer will be the same as the one obtained using the tangent line up to `4` decimal places (so our answer, which is to three decimal places, is the same as the actual value of `root(3)(27.3)`).

Hopefully this helps!